Question

Part A: A 3.00-L flask is filled with gaseous ammonia, NH3. The gas pressure measured at 25.0 ∘C is 1.15 atm . Assuming ideal gas behavior, how many grams of ammonia are in the flask?

Part B If 1.00 mol of argon is placed in a 0.500-L container at 20.0 ∘C , what is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation)?

For argon, a=1.345(L2⋅atm)/mol2 and b=0.03219L/mol.

Answer 1

Part A Following ideal gas below in the given equation as:

PV = nRT

P is the pressure of the gas = 1.15 atm

V is the volume of the gas = 3 L

n is the amount of substance of gas (also known as number of moles) = w / M = w / 17.03

T is the temperature of the gas = 25 + 273 = 298 K

R = 0.0821 L atm K^-1 mol^-1 if
Pressure is in atmospheres(atm)
Volume is in litres(L)
Temperature is in kelvin(K)

substituting all the values

1.15 x 3 = (w/17.03) x 0.082 x 298

w =  (1.15 x 3 x 17.03) / (0.082 x 298) = 58.7535 / 24.436 = 2.40 grams of NH3

PART B We will make use of both ideal gas and vander waals gas equation as follows

1. Following Idea gas equation

PV = nRT

P = (1 x 0.0821 x 293 ) / 0.500 = 48.11 atm

  2. Following Vander Waals Gas equation as follows

a = 1.345(L2⋅atm)/mol2 and b = 0.03219L/mol.

(P + n^2 a / V^2 ) (V - nb) = n R T

[ P + (1 x 1.345) / 0.5^2 ] x [ 0.5 - 1 x 0.03219] = 1 x 0.0821 x 293 ]

[ P + 1.345 ] x [ 0.45781 ] = 24.05

0.45781P + 0.61575 = 24.05

P = 51.18 atm

so the difference between the ideal pressure and real pressure is 51.18 - 48.11 = 3.07 atm